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Snubber RC

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Exemple Parasitics: C = 214.2 pF, L = 295.7 nH, ringing at 20 MHz with Z₀ = 37.15 Ω — fit R = 36 Ω (nearest E24 to 37.15 Ω) in series with C = 1 nF (first E12 above 4× the parasitic)

Exemple R burns 57.6 mW on average at 100 kHz, but every edge dumps 288 nJ into it in one go, at 666.7 mA peak (Vsw/R) — that pulse is small enough for an ordinary part, so a 0.125 W non-inductive resistor covers it (twice the average, because the heat still arrives in bursts). R·C = 36 ns against a 50 ns ring period, which is what "damped in about one cycle" looks like

Two-measurement method. You are giving the ring as the scope sees it, then again with a known capacitor clipped across the switch. $C_p=\frac{C_{\mathrm{add}}}{{\left(\frac{f_0}{f_1}\right)}^2-1}$ comes out of the two frequencies alone — the probe's own capacitance cancels — and $L_p=\frac{1}{{\left(2\pi f_0\right)}^2{C}_p}$ follows from the first reading.

$Z_0=\sqrt{\frac{L_p}{C_p}}$ is the ring's characteristic impedance — the number of ohms the tank is missing. Set $R_s=Z_0$ and the ringing is gone in about one cycle; $C_s=4C_p$ keeps the snubber capacitor large enough to dominate the parasitic it is fighting instead of joining it.

The price is heat: $P_R=C_s{V_{\mathrm{sw}}}^2{f}_{\mathrm{sw}}$ on average. But the average is not what the resistor has to survive — the whole $E_{\mathrm{edge}}=\frac{1}{2}{C}_s{V_{\mathrm{sw}}}^2$ lands in it in one shot on every edge, together with a $\frac{V_{\mathrm{sw}}}{R_s}$ spike, which is why the wattage below is only offered once that pulse is small enough for an ordinary part. The other classic mistake is reaching for a wirewound resistor, whose own inductance is exactly what the snubber is there to remove.